Question

In triangle abc , angle acb is an obtuse angle. Seg ad is perpendicular to line bc, show that ab square = bc sq + ac sq + 2bc.Cd

Answer 1

Answer:

The proof is explained step-wise below :

Step-by-step explanation:

Since, ∠ACB is given to be obtuse so the perpendicular AD will meet the BC by extending the line BD to point D.

For better explanation of the solution, see the attached figure of the diagram :

Sine, AD is the perpendicular so ΔABD is right angle triangle.

Now by Pythagoras theorem in ΔABD,

AB² =AD² + BD²

AB² = AD² + (BC+CD)²

AB² = AD² + BC² + CD² + 2×BC×CD

AB² = BC² + AD² + CD² + 2·BC×CD

AB² = BC² + AC² + 2·BC×CD

(since ADC is also right angle triangle using Pythagoras theorem in ΔADC

⇒ AC² = AD² + BD² )

⇒ AB² = BC² + AC² + 2·BC×CD

Hence Proved.

Answer 2

Answer:

Step-by-step explanation: